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Forces, Friction, and work

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A factory worker pushes a 28.4 kg crate a distance of 4.6 m along a level floor at constant velocity by pushing downward at an angle of 31 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

    What magnitude of force (Fpush) must the worker apply to move the crate at constant velocity?

    2. Relevant equations

    sum of Fx = Fpush*cos(31) - (kinetic friction) = 0
    sum of Fy = n - Fpush*sin(31) - mg = 0

    total work = K2 - K1 = 0
    work = F dot s = Fscos(31)

    kinetic friction = mu-sub-k * n

    3. The attempt at a solution

    Fpush = (kinetic friction) / cos(31)


    n = Fpush*sin(31) + mg

    I need n, but I can't find it this way, what else can I do?

    Thanks for any help.
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2
    Solve for Fpush and plug that into the other equation. I'm having trouble with forces myself, but that's what I would do.
  4. Feb 15, 2009 #3
    Actually FN is equal to mass times gravity. You already have everything else so I think all you have to do is solve.
  5. Feb 16, 2009 #4
    I tried keeping n as the mass times gravity, but I get the wrong answer since there is a downward force exerted on the crate by the factory worker.
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